Problem: Complete the square to solve for $x$. $x^{2}+13x+36 = 0$
Move the constant term to the right side of the equation. $x^2 + 13x = -36$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. The coefficient of our $x$ term is $13$ , so half of it would be $\dfrac{13}{2}$ , and squaring it gives us ${\dfrac{169}{4}}$ $x^2 + 13x { + \dfrac{169}{4}} = -36 { + \dfrac{169}{4}}$ We can now rewrite the left side of the equation as a squared term. $( x + \dfrac{13}{2} )^2 = \dfrac{25}{4}$ Take the square root of both sides. $x + \dfrac{13}{2} = \pm\dfrac{5}{2}$ Isolate $x$ to find the solution(s). $x = -\dfrac{13}{2}\pm\dfrac{5}{2}$ The solutions are: $x = -4 \text{ or } x = -9$ We already found the completed square: $( x + \dfrac{13}{2} )^2 = \dfrac{25}{4}$